Question

question 11 of 25
the price of 9 - volt batteries is increasing according to the function below, where t is years after january 1, 1980. during what year will the price reach $4?
$p(t)=1.1cdot e^{0.047t}$
a. 2005
b. 2009
c. 2003
d. 2007

Explanation:

Step1: Set up the equation

Set \(P(t) = 4\) in the formula \(P(t)=1.1\cdot e^{0.047t}\). So, \(4 = 1.1\cdot e^{0.047t}\).

Step2: Solve for \(e^{0.047t}\)

Divide both sides by \(1.1\): \(\frac{4}{1.1}=e^{0.047t}\), which simplifies to \(\frac{40}{11}=e^{0.047t}\).

Step3: Take the natural - logarithm of both sides

Using the property \(\ln(a^b)=b\ln(a)\) and \(\ln(e^x)=x\), we have \(\ln(\frac{40}{11})=\ln(e^{0.047t})\). So, \(\ln(\frac{40}{11}) = 0.047t\).

Step4: Calculate \(\ln(\frac{40}{11})\)

\(\ln(\frac{40}{11})\approx\ln(3.6364)\approx1.291\).

Step5: Solve for \(t\)

Since \(0.047t=\ln(\frac{40}{11})\approx1.291\), then \(t=\frac{1.291}{0.047}\approx27.47\).

Step6: Find the year

Since \(t\) is years after 1980, the year is \(1980 + 27.47\approx2007\).

Answer:

D. 2007